 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-3/14e/b/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

K (T) = \frac{1}{n _{B}} (\frac{2 π m _{e} k _{B} T}{h ^{2}})^{3/2} = \frac{2 π}{8 η ζ ( 3 )} (\frac{m _{e} c ^{2}}{k _{B} T})^{3/2} .

Using

n_{e} = n_{B} F

n_{j^{+}} = X_{j^{+}} n_{j}

, and

n_{j^{0}} = (1 - X_{j^{+}}) n_{j}

, the two Saha relations become the closed system

F \frac{X _{H^{+}}}{1 - X _{H^{+}}} = K (T) e^{- I_{H} / (k_{B} T)},

F \frac{X _{H e^{+}}}{1 - X _{H e^{+}}} = 2 K (T) e^{- I_{He} / (k_{B} T)},

where

F = (1 - Y_{p}) X_{H^{+}} + \frac{Y _{p}}{4} X_{H e^{+}} .

These are two equations for the two ionization fractions.

When

Y_{p} \to 0

F = X_{H^{+}}

, and the hydrogen equation reduces to

\frac{X _{H^{+}}^{2}}{1 - X _{H^{+}}} = \frac{1}{n _{B}} (\frac{2 π m _{e} k _{B} T}{h ^{2}})^{3/2} e^{- I_{H} / (k_{B} T)},

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